TITLE: Eigenvalues and Eigenvectors Diagonilization
QUESTION [5 upvotes]: Let $ A=\begin{bmatrix}
        -7 & -1  \\
        12 & 0 \\  \end{bmatrix} $ . Find a matrix $ P $ and a diagonal matrix $D$ such that $PDP^{-1} = A$.
Ok so the first thing I need to look for are my eigenvalues and eigenvectors. However, I think I'm doing it wrong and don't seem to get the correct eigenvectors. 
If $ det(A-\lambda I)X=0 $ then
$ det(\begin{bmatrix}
        -7-\lambda & -1  \\
        12 & -\lambda \\  \end{bmatrix})= \lambda^2+7\lambda+12=0 $
This gives me eigenvalues $\lambda_1=-3 $ and $\lambda_2 = -4 $

Case: $ \lambda_1 = -3$
$ \begin{bmatrix}
        -7-(-3) & -1  \\
        12 & -(-3) \\  \end{bmatrix} = \begin{bmatrix}
        -4 & -1  \\
        12 & 3 \\  \end{bmatrix} $ and end up with eigenvector
$\begin{bmatrix}
        -1/4  \\
        1 \\  \end{bmatrix} $ 
But this is wrong, and not sure where I made the error, or if my whole procedure is off. So I stopped here.

REPLY [2 votes]: You can write the solution as:
$$ A= P \cdot D \cdot P^{-1} = \begin{bmatrix} -1 & -1\\3 & 4\end{bmatrix} \cdot \begin{bmatrix} -4 & 0\\0 & -3 \end{bmatrix}  \cdot \begin{bmatrix} -4 & -1\\3 & 1\end{bmatrix}$$
Compare the eigenvalues\eigenvectors I wrote and the ones you wrote. So your answer is exactly the same as mine! Look at them closely and ask what you can change to make them look identical.
I would rather stay away from the fractions as they make my math more error prone (just a choice).
Does that make sense?